\newproblem{lay:2_1_26}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.26}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $A$ be an $m\times n$ matrix. Suppose there exists an $n\times m$ matrix $D$ such that $AD=I_m$ (the $m\times m$ identity matrix).
	Show that for any $\mathbf{b}\in\mathbb{R}^m$, the equation $A\mathbf{x}=\mathbf{b}$ has a solution. Explain
	why $A$ cannot have more rows than columns.
}{
  % Solution
	Let us consider the rows of $D=\begin{pmatrix}\mathbf{d}_1 & \mathbf{d}_2 & ... & \mathbf{d}_m\end{pmatrix}$. The product $AD$ is
	\begin{center}
		$AD=\begin{pmatrix}A\mathbf{d}_1 & A\mathbf{d}_2 & ... & A\mathbf{d}_m\end{pmatrix}=I_m=\begin{pmatrix}\mathbf{e}_1 & \mathbf{e}_2 & ... & \mathbf{e}_m\end{pmatrix}$
	\end{center}
	where $\mathbf{e}_i$ is the $i$-th column of $I_m$. For a particular column, we have
	\begin{center}
		$A\mathbf{d}_i=\mathbf{e}_i$
	\end{center}
	The columns of $I_m$ form a basis of $\mathbb{R}^m$. Therefore, for any $\mathbf{b}\in\mathbb{R}^m$ it can be expressed as a linear combination of the $\mathbf{e}_i$
	vectors
	\begin{center}
		$\mathbf{b}=\sum\limits_{i=1}^m{b_i\mathbf{e}_i}=\sum\limits_{i=1}^m{b_iA\mathbf{d}_i}=A\left(\sum\limits_{i=1}^m{b_i\mathbf{d}_i}\right)$
	\end{center}
	So we deduce, there exists a solution to the equation $A\mathbf{x}=\mathbf{b}$ that is
	\begin{center}
		$\mathbf{x}=\sum\limits_{i=1}^m{b_i\mathbf{d}_i}$
	\end{center}
	If $A$ had more rows than columns, then it would not have a solution for every $\mathbf{b}$ because there would be $\mathbf{b}$'s for which the reduced echelon form 
	has rows full of zeros and the independent terms are not 0.
}
\useproblem{lay:2_1_26}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
